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General relativity and newton's laws

  1. Jun 24, 2004 #1
    Is the gravitational force acting on a particle of mass m, on the
    surface of a sphere of radius 10^24 metres and with a mass of
    10^52 kg given by G x10^52 m / (10^24 ) ^ 1/2 - the Newtonian value - or is the mass density high enough for general relativity to be required to get a
    sensible result?
     
  2. jcsd
  3. Jun 24, 2004 #2

    DW

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    r is squared, not square rooted. In general relativity The law of motion can be written [tex]F^\lambda = m(\frac{dU^\lambda }{d\tau }) + m\Gamma ^{\lambda }_{\mu }_{\nu }U^{\mu }U^{\nu }[/tex].
    The real force on your test mass is the normal force holding it up and is expressed as the four vector on the left hand side of the equation. The last expression on the right can be called the gravitational force, but is only an inertial force which are sometimes called fictitious forces. What I think you really intend to compare is the prediction from Newtonian mechanics for the weight measured on a scale Vs the prediction for the measurement from general relativity. What the scale realy reads is the reaction force on it associated to the real normal force up on the test mass. The readout for the general relativisitc prediction would be given from equation 10.2.1 at
    http://www.geocities.com/zcphysicsms/chap10.htm#BM10_2
    [tex]F'_{felt} = \frac{GMm/r^{2}}{\sqrt{1 - \frac{2GM}{rc^2}}}[/tex]
     
    Last edited: Jun 24, 2004
  4. Jun 24, 2004 #3
    Apparently, a mass can't be held still over the event horizon of a black hole, so the mass I was talking about would have to be in motion on the surface of the sphere.
    If gravitons exist would they have to account for the "felt" force or F lambda?
     
  5. Jun 24, 2004 #4

    DW

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    Under. Over it can be held still just fine. Under the event horizon of a Shwarzschild hole all things are constrained to fall toward the physical singularity. Whatever matter you have producing the normal force up on the test mass is what is responsible for that force. General relativity is not a quantum theory and as such has no gravitons. Those are proposed by particle exchange theories.
     
    Last edited: Jun 24, 2004
  6. Jun 24, 2004 #5
    The gravitational field alters the mass of the particle. If the proper mass of the particle is m0, and the particle is not moving, then the mass of the particle is

    [tex]m = m_{0}\frac{dt}{d\tau} = \frac{ m_{0} }{ \sqrt{1 + 2\Phi/c^2}}[/tex]

    where

    [tex]\Phi = -\frac{GM}{r}[/tex]

    The gravitational force G is then given by

    [tex]G = \frac{GMm}{r^{2}}[/tex]

    Pete
     
    Last edited: Jun 24, 2004
  7. Jun 24, 2004 #6

    DW

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    Mass is invariant. Even in the presence of a gravitational field, Riemannian spacetime curvature, the quantity
    [tex]m = \frac{\sqrt{|g_{\mu }_{\nu }p^{\mu }p^{\nu }|}}{c}[/tex]
    which is defined as the mass of a free particle for modern general relativity does not depend on frame, speed, position, etc. Pmb's expression is not generally covariant-
    http://groups.google.com/groups?q=+...=off&selm=34EE0B44.399A3FA1@lucent.com&rnum=7
    and as such is not representative of modern general relativity. On a more basic note, it is bad to use the same case of the same letter to represent two different things in the same equation as in pmb's
    "[tex]G = \frac{GMm}{r^2}[/tex]"
     
    Last edited: Jul 14, 2004
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